Express x^2-14x+55 in the form (x-a)^2+b where you will find the values of the constants a and b. Then show that x^2-14x+55 is positive for all values of x.
2007-05-20 02:51:05 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2 - 14x + 49 = 6
(x - 7)^2 = 6
(x - 7)^2 - 6 = 0
x - 7 = +or- √6
x = 7 +or- √6
But going back to the equation in the form you requested (the third one), we know that all squares are positive. If you have to subtract 6 from x to get to 0, then x must be positive.
2007-05-20 03:12:46 · answer #1 · answered by Don E Knows 6 · 0⤊ 0⤋
This expression can be written as:
x^2-14x+49+6
Then
x^2-14x+49 = (x-7)^2
Then the expression is equal to
(x-7)^2+6
You can easily figure out that
a=7 and b=6
The square of any one number no matter whether it is positive or negative or zero is always positive or 0. When we add a positive number (6) to a 0 or another positive number the result is also positive.
2007-05-20 10:00:09 · answer #2 · answered by JAM-Just Ask Me :-) 2 · 0⤊ 0⤋
x^2 - 14x + 55 = x^2 - 14x + 49 + 6
= (x-7)^2 + 6
Obviously (x-7)^2 is always positive, and so it is +6
2007-05-20 09:58:25 · answer #3 · answered by gesges 3 · 0⤊ 0⤋
(x-7)^2 + 6 = x^2-14x+55
Since ANYTHING squared is always positive, and 6 is positive, the final value would be positive for any value of x
2007-05-20 11:14:01 · answer #4 · answered by Zilla 2 · 0⤊ 0⤋
y=(x-7)^2+6
(x-7)^2>=0 and 6>0 so the sum is>0 for all x
2007-05-20 09:59:07 · answer #5 · answered by santmann2002 7 · 0⤊ 0⤋